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Sagot :
SOLUTION:
Step 1: Calculate the molar mass of NaHCO₃.
Note that the molar masses of Na, H, C, and O are 22.99 g, 1.008 g, 12.011 g, and 15.999 g, respectively.
[tex]\begin{aligned} \text{molar mass of} \: \text{NaHCO}_3 & = \text{22.99 g + 1.008 g + 12.011 g + 3(15.999 g)} \\ & = \text{84.006 g} \end{aligned}[/tex]
Step 2: Solve for the percentage composition of NaHCO₃.
• For Na
[tex]\begin{aligned} \%\text{Na} & = \frac{n \times \text{molar mass of Na}}{\text{molar mass of} \: \text{NaHCO}_3} \times 100\% \\ & = \frac{1 \times \text{22.99 g}}{\text{84.006 g}} \times 100\% \\ & = \boxed{27.367\%} \end{aligned}[/tex]
• For H
[tex]\begin{aligned} \%\text{H} & = \frac{n \times \text{molar mass of H}}{\text{molar mass of} \: \text{NaHCO}_3} \times 100\% \\ & = \frac{1 \times \text{1.008 g}}{\text{84.006 g}} \times 100\% \\ & = \boxed{1.200\%} \end{aligned}[/tex]
• For C
[tex]\begin{aligned} \%\text{C} & = \frac{n \times \text{molar mass of C}}{\text{molar mass of} \: \text{NaHCO}_3} \times 100\% \\ & = \frac{1 \times \text{12.011 g}}{\text{84.006 g}} \times 100\% \\ & = \boxed{14.298\%} \end{aligned}[/tex]
• For O
[tex]\begin{aligned} \%\text{O} & = \frac{n \times \text{molar mass of O}}{\text{molar mass of} \: \text{NaHCO}_3} \times 100\% \\ & = \frac{3 \times \text{15.999 g}}{\text{84.006 g}} \times 100\% \\ & = \boxed{57.135\%} \end{aligned}[/tex]
Hence, NaHCO₃ is composed of 27.367% Na, 1.200% H, 14.298% C, and 57.135% O by mass.
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Note: The value of n in the formula of percent composition is the subscript of each element in the compound. If there is no subscript, it means that the value of n is 1.
[tex]\\[/tex]
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